Philip Wong
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hi guys,
I know this sound strange but it did happen. I was working on a question asking about calculating the resistance of a resistors, using different material one is copper and another is aluminium. I got the right answer, and wrote down the formula and steps that I used. But then I only didn't put in the values that I actually used. Just formula, steps and answer. Now when I come back to look at it again, I couldn't get the same answer that I originally correct. Either could I back track how I did it, so I hope someone can point it out to me, how I did it in the first place, and did I do it right then.
a 20mm long piece of copper wire (sigma = 5.8*10^7 (ohm.m)^-1) with a diameter of 0.01mm is used as a resistor. you want to design a resistor with the same resistance but the only wire avaiable to you is an aluminium wire (sigma = 3.4*10^7)ohm.m)^-1) with diameter 0.02 mm. calculate the length for the piece of aluminium wire for which the two resistances are the same
R = pL/A
the steps that I wrote down original listed below, and I haven't put in the fitted values then, but as I tries to put in all the values in again, I didn't get a consistence result.
Rcu = pL/A
= L/ (sigma* A)
= (L/sigma) (4/ pi d^2)
=4.39ohm
Values I used to reattempt to get answer:
L = 20mm = 0.02m
sigma = 5.8*10^7
d = (0.01*10-3) ^2
Rcu = (0.02/(5.8*10^7)) * (4/ (pi*(.01*10^-3)^2
Why...?
I know this sound strange but it did happen. I was working on a question asking about calculating the resistance of a resistors, using different material one is copper and another is aluminium. I got the right answer, and wrote down the formula and steps that I used. But then I only didn't put in the values that I actually used. Just formula, steps and answer. Now when I come back to look at it again, I couldn't get the same answer that I originally correct. Either could I back track how I did it, so I hope someone can point it out to me, how I did it in the first place, and did I do it right then.
Homework Statement
a 20mm long piece of copper wire (sigma = 5.8*10^7 (ohm.m)^-1) with a diameter of 0.01mm is used as a resistor. you want to design a resistor with the same resistance but the only wire avaiable to you is an aluminium wire (sigma = 3.4*10^7)ohm.m)^-1) with diameter 0.02 mm. calculate the length for the piece of aluminium wire for which the two resistances are the same
Homework Equations
R = pL/A
The Attempt at a Solution
the steps that I wrote down original listed below, and I haven't put in the fitted values then, but as I tries to put in all the values in again, I didn't get a consistence result.
Rcu = pL/A
= L/ (sigma* A)
= (L/sigma) (4/ pi d^2)
=4.39ohm
Values I used to reattempt to get answer:
L = 20mm = 0.02m
sigma = 5.8*10^7
d = (0.01*10-3) ^2
Rcu = (0.02/(5.8*10^7)) * (4/ (pi*(.01*10^-3)^2
Why...?